Question:
The integral $\int\left(\frac{x}{x \sin x+\cos x}\right)^{2} d x$ is equal to
(where $C$ is a constant of integration):
Correct Option: 1
Solution:
$\int \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$
$\because \frac{d}{d x}(x \sin x+\cos x)=x \cos x$
$=\int \frac{x \cos x}{(x \sin x+\cos x)^{2}}\left(\frac{x}{\cos x}\right) d x$
$=\frac{x}{\cos x}\left[\frac{-1}{x \sin x+\cos x}\right]$
$-\int \frac{x \sin x+\cos x}{\cos ^{2} x}\left[\frac{-1}{x \sin x+\cos x}\right] d x$
$=\frac{x}{\cos x}\left[\frac{-1}{x \sin x+\cos x}\right]+\int \sec ^{2} x d x$
$=\frac{-x \sec x}{x \sin x+\cos x}+\tan x+C$
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